Number of vertices in a given graph. Is there any way to find number of vertices in a given graph when radius and diameter of a graph
is known? I know the result where we have a found on number of edges i.e.,
$e\leq \frac{n(n-1)}{2}$, where $n$ is the number of vertices in a graph.
Kindly help. Any hint or clue is appreciated. Thanks for the help.
 A: There is no hope to upper bound the number of vertices or edges in a graph, given its diameter and radius (that are allowable, of course), as the following construction shows. Let $d$ be the desired diameter and $r$ the radius. We begin with two disjoint paths, $P_r$ and $P_d$, of length $r$ and $d$, respectively. We continue adding leafs to the middle of $P_d$ (creating a star in the middle of the path), which creates arbitrarily many vertices and edges in this graph, but doesn't change the diameter or radius.
We may get a lower bound with another construction. Given that a graph $G$ has radius $r$ and diameter $d$, there exists a path $P_d$ of length $d$ in that graph. Thus, $e(G) \geq d$, $n(G) \geq d+1$. You may do some extra work to incorporate the radius, but be careful as to whether the path $P_r$ that witnesses the radius intersects $P_d$.
A: 
Here is a proposition from Diestel's book. I hope it helps.
A: Thanks for all the provided answers. I just found an article which gives minimum number of vertices in a graph when diameter and radius of graph are known to us. Here is the link to the article. Thanks once again. 
http://www.sciencedirect.com/science/article/pii/0012365X73901167
The result is as follows:
Theorem: For all positive integers $m$ and $n$ satisfying $m\leq n\leq 2m-2$ there exist graphs of radius
$m$ and diameter $n$. The minimum order of such a graph is $n+m$.
